diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md
index 00caeea0..c7f79b8d 100644
--- a/problems/0019.删除链表的倒数第N个节点.md
+++ b/problems/0019.删除链表的倒数第N个节点.md
@@ -87,30 +87,31 @@ public:
 java:
 
 ```java
-class Solution {
-    public ListNode removeNthFromEnd(ListNode head, int n) {
-        ListNode dummy = new ListNode(-1);
-        dummy.next = head;
+public ListNode removeNthFromEnd(ListNode head, int n){
+    ListNode dummyNode = new ListNode(0);
+    dummyNode.next = head;
 
-        ListNode slow = dummy;
-        ListNode fast = dummy;
-        while (n-- > 0) {
-            fast = fast.next;
-        }
-        // 记住 待删除节点slow 的上一节点
-        ListNode prev = null;
-        while (fast != null) {
-            prev = slow;
-            slow = slow.next;
-            fast = fast.next;
-        }
-        // 上一节点的next指针绕过 待删除节点slow 直接指向slow的下一节点
-        prev.next = slow.next;
-        // 释放 待删除节点slow 的next指针, 这句删掉也能AC
-        slow.next = null;
+    //排除示例 2 的情况：head = [1], n = 1
+    if (head == null)
+        return null;
 
-        return dummy.next;
+    ListNode fastIndex = dummyNode;
+    ListNode slowIndex = dummyNode;
+
+    //只要快慢指针相差 n 个结点即可
+    for (int i = 0; i < n  ; i++){
+        fastIndex = fastIndex.next;
     }
+
+    while (fastIndex.next != null){
+        fastIndex = fastIndex.next;
+        slowIndex = slowIndex.next;
+    }
+
+    //此时 slowIndex 的位置就是待删除元素的前一个位置。
+    //具体情况可自己画一个链表长度为 3 的图来模拟代码来理解
+    slowIndex.next = slowIndex.next.next;
+    return dummyNode.next;
 }
 ```